第12类RD Sharma解决方案–第11章求差–练习11.4 |套装2

在以下每个中找到dy / dx。

问题16.如果x√(1 + y)+y√(1 + x)= 0，则证明(1 + x) ² dy / dx + 1 = 0

解决方案：

We have,

 x√(1+y) +y√(1+x) =0

=>x√(1+y)=-y√(1+x)

On squaring both sides, we have

x²(1+y)=y²(1+x)

=>x²+ x²y-y²-y²x=0

=>(x+y)(x-y)+xy(x-y)=0

=>(x-y)(x+y+xy)=0

So, either (x-y)=0

or, x+y+xy=0

=>x+y(1+x)=0

=>y=-x/(1+x)

On differentiating both sides with respect to x,

dy/dx=d(-x/(1+x))/dx

On applying quotient rule,

dy/dx = ((x)1-(1+x))/(1+x)²

=>dy/dx=(-1/(1+x)²)

=>(dy/dx)(1+x)²+1=0

Hence, proved.

问题17. log(√(x ² + y ² ))= tan ^-1 (y / x)

解决方案：

We have, 

log(√(x²+y²))=tan^-1(y/x)

=>log(x²+y²)^(1/2)=tan^-1(y/x)

=>(1/2)(log(x²+y²))=tan^-1(y/x)

On differentiating both sides with respect to x,

(1/2)d(log(x²+y²))/dx = d(tan^-1(y/x))/dx

=>(1/2)*(1/(x²+y²))*(2x+2y(dy/dx))=1/(1+(y/x)²)((x(dy/dx)-y)/x²)

=>x+y(dy/dx)=x(dy/dx)-y

=>(dy/dx)(y-x)=-(x+y)

=>(dy/dx)=(x+y/(x-y)

Therefore, the answer is,

(dy/dx)=(x+y)/(x-y)

问题18. sec((x + y)/(xy))= a

解决方案：

We have,

sec((x+y)/(x-y))=a

=>(x+y)/(x-y)=sec^-1(a)

On differentiating both sides with respect to x,

=>d((x+y)/(x-y))/dx=d(sec^-1(a))/dx

=>(x-y)(1+(dy/dx))-(x+y)(1-(dy/dx))=0(x-y)²

=>x-y+(x-y)(dy/dx)-(x+y)+(x+y)(dy/dx)=0

=>(dy/dx)(x-y+x+y)-2y=0

=>(dy/dx)(2x)=2y

=>(dy/dx)=(y/x)

Therefore, the answer is,

(dy/dx)=(y/x)

问题19.tan ^-1 ((x ² -y ² )/(x ² + y ² ))= a

解决方案：

We have,

tan^-1((x²-y²)/(x²+y²))=a

=>(x²-y²)/(x²+y²)=tan a

=>(x²-y²)=(tan a)(x²+y²)

On differentiating both sides with respect to x,

=>d(x²-y²)dx=d((tan a)(x²+y²))/dx

=>2x-2y(dy/dx)=(tan a)(2x+2y(dy/dx))

=>x-y(dy/dx)=(tan a)(x+y(dy/dx))

=>-(dy/dx)(y+y(tan a))=x(tan a)-x

=>-(dy/dx)=(x(tan a-1))/(y(1+tan a))

=>dy/dx= (x(1-tan a))/(y(1+tan a))

Therefore, the answer is,

dy/dx =(x(1-tana))/(y(1+tan a))

问题20. xy(log(x + y))= 1

解决方案：

We have,

xy(log(x+y))=1

Differentiating it with respect to x,

d(xy(log(x+y)))/dx =d1/dx

=>y(log(x+y))+x(log(x+y)dy/dx+((xy)/(x+y))(1+(dy/dx)))=0

=>y(log(x+y))+((xy)/(x+y))+(dy/dx)(x(log(x+y))+(xy)/(x+y))=0

=>(dy/dx)(x(log(x+y))+(xy)/(x+y))=-(y(log(x+y))+(xy)/(x+y))

It can be deduced that ,

y(log(x+y))=1/x

x(log(x+y))=1/y

So,

(dy/dx)((1/y)+(xy)/(x+y))=-((1/x)+(xy)/(x+y)

=>(dy/dx)((x+y+xy²)/((y+y)x))=-(x+y+x²y)/(y)(x+y))

=>(dy/dx)=-((x+y+x²y)/(x+y+xy²))(y/x)

Therefore, the answer is,

dy/dx=-((x(x²y+x+y))/(y(xy²+x+y)))

问题21. y = xsin(a + y)

解决方案：

We have,

y=x sin(a+y)

Differentiating it with respect to x,

dy/dx=sin(a+y) +x*cos(a+y){0+dy/dx}

=>dy/dx =sin(a+y) +x*cos(a+y)*(dy/dx)

=>dy/dx-x*cos(a+y)*(dy/dx) =sin(a+y)

=>(dy/dx)(1-x*cos(a+y))=sin(a+y)

=>dy/dx=(sin(a+y))/(1-x*cos(a+y))

Therefore, the answer is,

dy/dx =(sin(a+y))/(1-x*cos(a+y))

问题22. x * sin(a + y)+(sin a)*(cos(a + y))= 0

解决方案：

We have,

x*sin(a+y)+(sin a)*(cos(a+y))=0

On differentiating both sides with respect to x,

d(x*sin(a+y)+(sin a)*(cos(a+y)))/dx=d0/dx

=>sin(a+y)+x*cos(a+y)*(dy/dx)-(sin a)sin(a+y)(dy/dx)=0

=>(dy/dx)(xcos(a+y)-sina(sin(a+y)))=-sin(a+y)

=>(dy/dx)=sin(a+y)/(sina*sin(a+y)-xcos(a+y))

From above,

x=-((sina)*cos(a+y))/sin(a+y)

Putting in the above equation,

(dy/dx)*(((sina)*cos²(a+y))/(sin(a+y)))+(sina)sin(a+y))=sin(a+y)

(dy/dx)((sina)((cos²(a+y)+sin²(a+y))/sin(a+y)) =sin(a+y)

(dy/dx)=(sin²(a+y))/(sin a)

Therefore, the answer is,

(dy/dx)=sin²(a+y)/(sina)

问题23. y = x * siny

解决方案：

We have,

y=x*siny

On differentiating both sides with respect to x,

dy/dx=siny+x(cosy)(dy/dx)

=>dy/dx-x(cosy)(dy/dx)=siny

=>(dy/dx)(1-x(cosy))=siny

=>dy/dx=(siny)/(1-x(cosy))

Therefore, the answer is,

(dy/dx)=(siny)/(1-x(cosy))

问题24. y(x ² +1) ^1/2 = log((x ² +1) ^1/2 -x)

解决方案：

We have,

y(x²+1)^1/2=log((x²+1)^1/2-x)

Differentiating it with respect to x,

d(y(x²+1)^1/2)/dx=(((x²+1)^1/2-x)^-1/2)(2(x²+1))^-1/2(2x-1)

=>2xy(2(x²+1)^-1/2)+(x²+1)^1/2(dy/dx)=(((x²+1)^1/2-x)^-1/2)(x-(x²+1)^1/2)(x²+1))^-1/2

=>(dy/dx)(x²+1)^1/2=((((x²+1)^1/2-x)^-1/2)(x-(x²+1))^-1/2x)/(x²+1))-(xy)(x²+1)^-1/2

=>(dy/dx)(x²+1)^1/2=(-1/(x²+1)^1/2)-(xy)(x²+1)^-1/2

=>(dy/dx)(x²+1)^1/2=(-1-xy)(x²+1)^-1/2

=>(dy/dx)(x²+1)=-(1+xy)

=>(dy/dx)(x²+1)+xy+1=0

Therefore, the answer is,

(dy/dx)(x²+1)+xy+1=0

问题25. y =(log _cosx sinx)(log _sinx cosx) ^-1 + sin ^-1 (2x /(1 + x ² ))

在x = pi / 4处找到dy / dx

解决方案：

We have,

y=(log_cosxsinx)(log_sinxcosx)^-1+sin^-1(2x/(1+x²))

=>y=(log_cosxsinx)(log_cosxsinx)+sin^-1(2x/(1+x²))

=>y=(log_cosxsinx)²+sin^-1(2x/(1+x²))

=>y=((log sinx)/log(cosx))²+sin^-1(2x/(1+x²))

Differentiating it with respect to x,

dy/dx=d((log sin x)/log(cos x))²/dx+ d(sin^-1(2x/(1+x²)))

=>dy/dx =2((log sinx)/log(cosx))d((log sinx)/(log cosx))/dx+1/(√1-((2x)/(1+x²))²d(2x/(1+x²))/dx

=>dy/dx=2((log sinx)/log(cosx))*((log cosx)(1/sin x)*(cos x)-(log sinx)*(1/cosx)*(-sinx))/(log(cosx))² +((1+x²)/√(1+x⁴-2x²))((1+x²)2-4x²)/(1+x²)²

=>dy/dx=2(log sinx)/(logcosx)*((log cosx)*cotx+(log sinx)tanx)+((1+x²)/√(1-x²)²)(1-2x²)/(1+x²)²

=>dy/dx=(2(log sinx)*((log cosx)cotx+(log sinx)tanx))/(log cosx)³+2/(1+x²)

At x=pi/4

dy/dx=(2(log sin(pi/4))*((log(cos(pi/4) cot(pi/4)+(log sin(pi/4))tan(pi/4))/(log cos(pi/4))³+2/(1+(pi²)/16)

=>dy/dx=2(log(1/√2))*(log(1/√2)+log(1/√2))/(log(1/√2))³+32/(16+(pi)²)

=>dy/dx=4(1/(log(1/√2)))+32/(16+(pi)²)

=>dy/dx=4(1/((-1/2)log2)+32/(16+(pi)²)

=>dy/dx=32/(16+(pi)²)-8(1/log2)

 Therefore, the answer is,

(dy/dx)=32/(16+(pi)²)-8(1/log2)

问题26. sin(xy)+ y / x = x ² -y ²

解决方案：

We have,

sin(xy)+y/x=x²-y²

Differentiating it with respect to x,

d(sin(xy)+d(y/x))/dx =d(x²)/dx -d(y²)/dx

=>cos(xy)(x(dy/dx)+y) +(x(dy/dx)-y)(x^-2)=2x-2y(dy/dx)

=>x*cos(xy)(dy/dx) + ycos(xy)+(x^-1)(dy/dx)-y(x^-2)+2y(dy/dx)=2x

=>(dy/dx)(x*cos(xy)+x^-1+2y)=2x-ycos(xy)-y(x^-2)

=>dy/dx=(2x-ycos(xy)-y(x^-2))/(x*cos(xy)+x^-1+2y)

 Therefore, the answer is,

(dy/dx)=(2x-ycos(xy)-y(x^-2))/(x*cos(xy)+x^-1+2y)

问题27.(y + x) ^1/2 +(yx) ^1/2 = c

解决方案：

We have,

(y+x)^1/2+(y-x)^1/2=c

Differentiating it with respect to x,

(1/2)(y+x)^-1/2((dy/dx)+1) + (1/2)(y-x)^-1/2((dy/dx)-1)=0

=>(dy/dx)((1/2)(y+x)^-1/2+(1/2)(y-x)^-1/2) +(1/2)(y+x)^-1/2-(1/2)(y-x)^-1/2=0

=>(dy/dx)=((y-x)^-1/2-(y+x)^-1/2)/((y+x)^-1/2+(y-x)^-1/2)

By rationalisation of denominator,

(dy/dx)=(y+x)+(y-x)-2(y+x)^1/2(y-x)^1/2

=>dy/dx=(2y-2(y+x)^1/2(y-x)^1/2)/(x+y-y+x)

=>dy/dx=(y-((y²-x²)^1/2)/(x)

 Therefore, the answer is,

dy/dx=(y-((y²-x²)^1/2)/(x)

问题28. tan(x + y)+ tan(xy)= 1

解决方案：

We have,

tan(x+y)+tan(x-y)=1

Differentiating it with respect to x,

d(tan(x+y)+tan(x-y))/dx=d1/dx

=>sec²(x+y)(d(x+y)/dx)+sec²(x-y)d(x-y)/dx=0

=>sec²(x+y)(1+dy/dx)+sec²(x-y)(1-dy/dx)=0

=>(dy/dx)(sec²(x+y)-sec²(x-y))+sec²(x+y)+sec²(x-y)=0

=>dy/dx=(sec²(x+y)+sec²(x-y))/(sec²(x-y)-sec²(x+y))

Therefore, the answer is,

dy/dx=(sec²(x+y)+sec²(x-y))/(sec²(x-y)-sec²(x+y))

问题29. e ^x + e ^y = e ^{x + y}

解决方案：

We have,

d(e^x+e^y)/dx=de^(x+y)/dx

=>e^x+e^y(dy/dx)=e^(x+y)(1+(dy/dx))

=>(dy/dx)(e^y-e^(x+y))=e^(x+y)-e^x

=>(dy/dx)=(e^(x+y)-e^x)/(e^y-e^(x+y))

=>dy/dx=e^x(e^y-1)/e^y(1-e^x)

Therefore, the answer is,

dy/dx=e^x(e^y-1)/e^y(1-e^x)

问题30.如果cosy = xcos(a + y)。然后证明dy / dx =(cos ² (a + y))/ sin a

解决方案：

We have,

cosy=x*cos(a+y)

Differentiating it with respect to x,

d(cosy)/dx=d(x*cos(a+y))/dx

=>-siny(dy/dx)=cos(a+y)-xsin(a+y)(dy/dx)

=>xsin(a+y)(dy/dx)-siny(dy/dx)=cos(a+y)

=>(dy/dx)(xsin(a+y)-siny)=cos(a+y)

=>dy/dx=(cos(a+y))/(x*sin(a+y)-siny)

Also, x=cosy/cos(a+y)

Substituting it in the earlier statement,

(dy/dx)=(cos(a+y))/((cosy)sin(a+y)/cos(a+y))-siny)

=>dy/dx=cos²(a+y)/(cosy*sin(a+y)-siny(cos(a+y)))

=>dy/dx=cos²(a+y)/(sin(a+y-y))

=>dy/dx=cos²(a+y)/sin(a)

Therefore, the answer is,

dy/dx=cos²(a+y)/sin a